Search Results

As noted in the comment by user44191, one needs only check this for prime powers. Note that this is trivial for $q=2$ and so we may assume that we have an odd prime $q$. Then the claim is that when $k …

The Shapiro inequality is the statement that if $x_1, x_2, \dots, x_n$ are positive, with $x_{n+1}=x_1, x_{n+2}=x_2$, then $$\sum_{i=1}^{n} \frac{x_i}{x_{i+1}+x_{i+2}} \geq \frac{n}{2}.$$ This can be …

If I'm understanding your problem correctly, your sum diverges and does so very rapidly. Every positive integer of the form $2^7p$ for an odd prime $p$ is a rad-refactorable number, so your sum is bou …

Let $\sigma(n)$ be the sum of the divisors of $n$. Is it always true that if $n$ is odd, that $$n\mid\sum_{k=1}^{\frac{n-1}{2}}k^2\sigma(k)\sigma(n-k)?$$ I have checked this up to $n=100$, and I suspe …

This should be provable by standard although laborious methods. What follows is a proof sketch (I have not checked all the computational details but this method should work). We recall a few basic fac …

Conjecture 2 seems to be true. If $n \geq 2$ then $$\frac{1}{2}(k^2+k+2)k^{n-k-1}-1 \leq \left\{{n \atop k}\right\} \leq \frac{1}{2}{n \choose k} k^{n-k} < 2^n k^{n-k}. $$ (Inequalities from Here.) …

As far as I'm aware, we don't have any substantially non-trivial bounds on the behavior of $\psi(n)$ when $n$ is an odd perfect number. We can at least prove the following but none of these are diffi …

In general, very few prime factors in an odd perfect number can be of the form $n^2+1$. In particular, if N is an odd perfect number then $\frac{\sigma(N)}{N}=2$, and for any $m$ (perfect or not), $\f …

Heuristically this should be the case. For any prime p greater than 5, consider the set of numbers of the form $2^a3^b5^c p \pm 1$. The "probability" that one of of these is prime should be about $$\ …